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Let $f : \mathbb{R} \to \mathbb{R}$ be continuous, everywhere-differentiable, and bounded above by $u$. Let $S_k = \{x | f(x) \le k \}$ (thus, $S_u = \mathbb{R}^n$). Imagine sliding $k$ down continuously from $u$. Then $S_k$ would lose path-connectedness right at the moment that the line $f(x) = k$ becomes tangent to the global maximum $g$ of $f(x)$, since any further decrease in $k$ will cause $g \notin S_k$, and poking a hole in $\mathbb{R}$ causes it to lose path-connectedness.

What is the equivalent condition if we change the domain of $f$ from $\mathbb{R}$ to $\mathbb{R}^n$?

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If you consider a real-valued function like $arctan(x)$ that is increasing but bounded, the sets $S_k$ remain path-connected for all $k$. –  hardmath Oct 9 '12 at 23:55

2 Answers 2

I don't think there's an equivalent condition. Consider $f:{\Bbb R}^2\to{\Bbb R}:(x,y)\mapsto -x^2-y^2$, which is bounded above, but for which the corresponding $S_k$ would be connected for any $k$.

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Sure, but that doesn't mean that no condition exists - it just means that the condition we find should be unmeetable in your example function. –  GMB Oct 9 '12 at 23:59
    
Then you may want to clarify what you meant by "equivalent condition". We're a very precise bunch of people around here. –  user22805 Oct 10 '12 at 0:11

While there's no nice equivalent condition for the set losing path connectedness, there is a nice equivalent statement:

If $f\colon \mathbb{R}^2 \rightarrow \mathbb{R}$ has a unique global maximum $g$, then the set $S_k$ will lose simple connectedness right at $k=g$.

In many ways, simple connectedness is the one-dimension-higher analogue of path-connectedness. This generalizes to $n$-connectedness, which lets you take the above statement to even higher dimensions.

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